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\title{\heiti\zihao{2} 习题15.5}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{求下列曲线在给定点处的切线与法平面:}
\subsection{$x=a \sin ^{2} t, y=b \sin t \cos t, z=c \cos ^{2} t$, 在点 $t=\dfrac{\pi}{4}$.}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\mathrm{d}x}{\mathrm{d}t} & =2a\sin t\cos t        \\
		\dfrac{\mathrm{d}y}{\mathrm{d}t} & =b(\cos^2 t - \sin^2t) \\
		\dfrac{\mathrm{d}z}{\mathrm{d}t} & =-2c\sin t\cos t       \\
	\end{aligned}
$$
切线:
$$
	\dfrac{x-\dfrac{a}{2}}{a}=\dfrac{y-\dfrac{b}{2}}{0}=\dfrac{z-\dfrac{c}{2}}{-c}
$$
法平面:
$$
	a\left(x-\dfrac{a}{2}\right)-c\left(z-\dfrac{c}{2}\right)=0
$$
\subsection{$2 x^{2}+3 y^{2}+z^{2}=9, z^{2}=3 x^{2}+y^{2}$,在点 $(1,-1,2)$.}
\textbf{解}\quad
的Jacobian矩阵为$\left[\begin{array}{ccc}
			4x & 6y & 2z  \\
			6x & 2y & -2z
		\end{array}\right]$.法向量为
$$
	\left(\left|\begin{array}{cc}
			\dfrac{\partial F_1}{\partial y} & \dfrac{\partial F_1}{\partial z} \\
			\dfrac{\partial F_2}{\partial y} & \dfrac{\partial F_2}{\partial z} \\
		\end{array}\right|,\left|\begin{array}{cc}
			\dfrac{\partial F_1}{\partial z} & \dfrac{\partial F_1}{\partial x} \\
			\dfrac{\partial F_2}{\partial z} & \dfrac{\partial F_2}{\partial x} \\
		\end{array}\right|,\left|\begin{array}{cc}
			\dfrac{\partial F_1}{\partial x} & \dfrac{\partial F_1}{\partial y} \\
			\dfrac{\partial F_2}{\partial x} & \dfrac{\partial F_2}{\partial y} \\
		\end{array}\right|\right)=(8,10,7)
$$
所以切线为
$$
	\dfrac{x-1}{8}=\dfrac{y+1}{10}=\dfrac{z-2}{7}
$$
法平面为
$$
	8(x-1)+10(y+1)+7(z-2)=0
$$

\section{求过直线 $L:\left\{\begin{array}{l}x-y+z=0 \\ x+2 y+z=1\end{array}\right.$ 与曲面 $\Sigma: x^{2}+y^{2}-z^{2}=1$ 相切的平面的方程.}
\textbf{解}\quad
设切点$(x_0,y_0,z_0)$,该点处法向量:$\vec{n}=(2x_0,2y_0,2z_0)$.其满足$x_0^2+y_0^2-z_0^2=0$.又有过$L$的平面系:$x-y+z+\lambda(x+2y+z-1)=0$,所以联立解得$k=1$或$k=\dfrac{1}{4}$.\par
检验:平面$x+2y+z=1$不于$\vec{n}$垂直,所以不满足要求.
故平面方程为$2x+y+2z-1=0$或$\dfrac{5}{4}x-\dfrac{1}{2}y+\dfrac{5}{4}z-\dfrac{1}{4}=0$.

\section{求曲线 $x=t, y=t^{2}, z=t^{3}$ 上一点,使曲线在此点的切线平行于平面 $x+2 y+z=4 .$}
\textbf{解}\quad
即需要让切线垂直于$\vec{n}=(1,2,1)$.\par
切向量:$(1,2t,3t^2)$.有$1+4t+3t^2=0$,解得$t=-1$或$t=-\dfrac{1}{3}$.\par
检验:由于$(1,1,1)$在平面上,故舍去.所以综上,该点为$\left(-\dfrac{1}{3},\dfrac{1}{9},-\dfrac{1}{27}\right)$.

\section{求曲线 $x=\dfrac{1}{4} t^{4}, y=\dfrac{1}{3} t^{3}, z=\dfrac{1}{2} t^{2}$ 的平行于平面 $x+3 y+2 z=0$ 的切线方程.}
\textbf{解}\quad
即垂直于$\vec{n}=(1,3,2)$的切线方程.\par
切向量:$(t^3,t^2,t)$,有$t^3+3t^2+t=0$,解得$t=0$或$t=-1$或$t=-2$.由于$t=0$时的点在平面上故舍去.\par
点$\left(\dfrac{1}{4},-\dfrac{1}{3},\dfrac{1}{2}\right)$处,切线方程为$\dfrac{x-\dfrac{1}{4}}{-1}=\dfrac{y+\dfrac{1}{3}}{1}=\dfrac{z-\dfrac{1}{2}}{-1}$\par
点$\left(4,-\dfrac{8}{3},2\right)$处,切线方程为$\dfrac{x-4}{-8}=\dfrac{y+\dfrac{8}{3}}{4}=\dfrac{z-2}{-2}$


\section{设函数 $F(u, v)$ 可微,试证: 曲面 $F\left(\dfrac{x-a}{z-c}, \dfrac{y-b}{z-c}\right)=0$ 上任意点处的切平面都通过一定点.}
\begin{proof}
	在曲面上的任意一点$(x_0,y_0,z_0)$,因为该点的法向量为
	$$
		\left(\dfrac{F_1}{z-c},\dfrac{F_2}{z-c},\dfrac{a-x}{(z-c)^2}F_1+\dfrac{b-y}{(z-c)^2}F_2\right)
	$$
	故切平面为
	$$
		\dfrac{F_1}{z-c}(x-x_0)+\dfrac{F_2}{z-c}(y-y_0)+\left(\dfrac{a-x}{(z-c)^2}F_1+\dfrac{b-y}{(z-c)^2}F_2\right)(z-z_0)=0
	$$
	解得$x=a,y=b,z=c$.所以平面束过点$(a,b,c)$.
\end{proof}

\section{求下列曲面在给定点处的切平面与法线:}
\subsection{$y-\mathrm{e}^{2 x-z}=0$, 在点 $(1,1,2)$}
\textbf{解}\quad
法向量:$\vec{n}=(-2e^{2x-z},1,e^{2x-z})$.在点$(1,1,2)$处:\par
切平面:$-2(x-1)+(y-1)+(z-2)=0$.\par
法线:$\dfrac{x-1}{-2}=\dfrac{y-1}{1}=\dfrac{z-2}{1}$

\subsection{$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}=1$, 在点 $\left(\dfrac{a}{\sqrt{3}}, \dfrac{b}{\sqrt{3}}, \dfrac{c}{\sqrt{3}}\right)$}
\textbf{解}\quad
法向量:$\vec{n}=(\dfrac{2x}{a^2},\dfrac{2y}{b^2},\dfrac{2z}{c^2})$.在点$(\dfrac{a}{\sqrt{3}},\dfrac{b}{\sqrt{3}},\dfrac{c}{\sqrt{3}})$处:\par
切平面:$\dfrac{2}{\sqrt{3}a}\left(x-\dfrac{a}{\sqrt{3}}\right)+\dfrac{2}{\sqrt{3}b}\left(y-\dfrac{b}{\sqrt{3}}\right)+\dfrac{2}{\sqrt{3}c}\left(z-\dfrac{c}{\sqrt{3}}\right)=0$.\par
法线:$a\left(x-\dfrac{a}{\sqrt{3}}\right)=b\left(y-\dfrac{b}{\sqrt{3}}\right)=c\left(z-\dfrac{c}{\sqrt{3}}\right)$.

\section{求曲面 $x^{2}+2 y^{2}+3 z^{2}=21$ 的切平面,使它平行于平面 $x+4 y+6 z=0$}
\textbf{解}\quad
即其与$\vec{n}=(1,4,6)$垂直.\par 
设在点$(x_0,y_0,z_0)$处的切平面与$\vec{n}$垂直,有该点的法向量:$(2x_0,4y_0,6z_0)$与$\vec{n}$共线.所以解得$2x_0=y_0=z_0$.又因为$x_0^2+2y_0^2+3z_0^2-21=0$,解得$(x_0,y_0,z_0)=(1,2,2)$或$(x_0,y_0,z_0)=(-1,-2,-2)$.\par 
点$(1,2,2)$处切平面:$x+4y+6z=21$\par
点$(-1,-2,-2)$处切平面:$x+4y+6z=-21$
\end{document}